Surface Area of a Right Circular Cylinder and Right Circular Cone, Class 9, Mathematics | Extra Documents & Tests for Class 9 PDF Download

RIGHT CIRCULAR CYLINDER
Solids like circular pillars, circular pipes, circular pencils, measuring jars, road rollers and gas cylinders, etc., are said to be in cylindrical shape.

In mathematical terms, a right circular cylinder is a solid generated by the revolution of a rectangle about its sides.
Let the rectangle ABCD revolve about its side AB, so as to describe a right circular cylinder as shown in the figure.
You must have observed that the cross-sections of a right circular cylinder are circles congruent and parallel to each other.

Cylinders Not Right Circular
There are two cases when the cylinder is not a right circular cylinder.

REMARK : Unless stated otherwise, here in this chapter the word cylinder would mean a right circular cylinder.
The following are definitions of some terms related to a right circular cylinder :

(i) The radius of any circular end is called the radius of the right circular cylinder.
Thus, in the above figure, AD as well as BC is a radius of the cylinder.

(ii) The line joining the centres of circular ends of the cylinder, is called the axis of the right circular cylinder.
In the above figure, the line AB is the axis of the cylinder. Clearly, the axis is perpendicular to the circular ends.

REMARK : If the line joining the centres of circular ends of a cylinder is not perpendicular to the circular ends, then the cylinder is not a right circular cylinder.

(iii) The length of the axis of the cylinder is called the height or length of the cylinder.

(iv) The curved surface joining the two bases of a right circular cylinder is called its lateral surface.
For a right circular cylinder of radius = r units & height = h units, we have :

The above formulae are applicable to solid cylinders only.

HOLLOW RIGHT CIRCULAR CYLINDERS
Solids like iron pipes, rubber tubes, etc., are in the shape of hollow cylinders.
A solid bounded by two coaxial cylinders of the same height and different radii is called a hollow cylinder.


For a hollow cylinder of height h and with external and internal radii R and r respectively, we have :

Ex 5. A cylindrical block of wood has radius 70 cm and length 2 m is to be painted with blue coloured enamel. The cost of painting is Rs. 1.25 per 100 cm². Find the cost of painting the block. Take π = 22/7 .
 Sol. Here, the radius r of the cylindrical block of wood = 70 cm and the length h = 200 cm
The total surface area of the cylindrical block = 2πr(r + h)
 cm2 = 440 × 270 cm² = 118800 cm²
Cost of painting 100 cm² = Rs. 1.25 =

⇒ Cost of painting 1 cm2 = Rs. (5over 400)

Then, cost of painting 118800 cm² = Rs. (5over 400) × 118800 =  × 1188 = Rs.1485
Hence, cost of painting the block of wood = Rs.1485.

Ex 6. A cylindrical vessel, without lid, has to be tin-coated including both of its sides. If the radius of its base is (1over 2) m and its height is 1.4 m, calculate the cost of tin-coating at the rate of Rs. 50 per 1000 cm². (Use π = 3.14)

Sol. Radius of the base (r) = (1over 2)m = (1over 2)× 100 cm = 50 cm
Height (h) = 1.4 m = 1.4 × 100 cm = 140 cm
Surface area to be tin-coated = 2 (2πrh + πr²)

 Cost of tin-coating at the rate of Rs. 50 per 1000 cm² =
Hence, the cost of tin-coating is Rs. 5181.

 Ex 7. The pillars of a temple are cylindrical shaped. If each pillar has a circular base of radius 20 cm and height 7 m, then find the quantity of concrete mixture used to build 20 such pillars. Also find the cost of the concrete mixture at the rate of Rs. 200 per m³. Take π  = (22over 7)

Sol. Concrete mixture used for making each pillar = Volume of each pillar = π r² × h

Hence, volume of mixture required to make 20 pillars = 17.6 m³
Now, cost of mixture at the rate of Rs.200 per m³
= Rs.200 × 17.6 = Rs. 3520.
Hence, the cost of concrete mixture is Rs. 3520.

RIGHT CIRCULAR CONE
Solids like an ice-cream cone, a conical tent, a conical vessel, a clown's cap etc. are said to be in conical shape. In mathematical terms, a right circular cone is a solid generated by revolving a right-angled triangle about one of the sides containing the right angle.
Let a triangle AOC revolve about it's side OC, so as to describe a right circular cone, as shown in the figure.

Cones Not Right Circular

REMARK : Unless stated otherwise, by 'cone' in this chapter, we shall mean 'a right circular cone'.
The following are definitions of some terms related to right circular cone :

(i) The fixed point O is called the vertex of the cone.
(ii) The fixed line OC is called the axis of the cone.
(iii) A right circular cone has a plane end, which is in circular shape. This is called the base of the cone. The vertex of a right circular cone is farthest from its base.
(iv) The length of the line segment joining the vertex to the centre of the base is called the height of the cone.

Length OC is the height of the cone.

(v) The length of the line segment joining the vertex to any point on the circular edge of the base, is called the slant height of the cone.
Length OA is slant height of the cone.

(vi) The radius AC of the base circle is called the radius of the cone.

 Relation Between Slant Height, Radius and Vertical Height.

Let us take a right circular cone with vertex at O, vertical height h, slant height  and radius r. A is any point on the rim of the base of the cone and C is the centre of the base. Here, OC = h, AC = r and OA = .

The cone is right circular and therefore, OC is at right angle to the base of the cone. So, we have OC  CA,
i.e., ΔOCA is right angled at C.
Then by Pythagoras theorem, we have :

For a right circular cone of Radius = r, Height = h & Slant Height = , we have :

Hollow Right Circular Cone

(i) Centre of the circle is vertex of the cone.
 (ii) Radius of the circle is slant height of the cone.
 (iii) Length of arc AB is the circumference of the base of the cone.
 (iv) Area of the sector is the curved surface area of the cone.

Ex 8. The radius of the base of a conical tent is 12 m. The tent is 9 m high. Find the cost of the canvas required to make the tent, if one square metre of canvas costs Rs. 120. (Take π = 3.14)
Sol. Here, r = 12 m and h = 9 m. Let  be slant height of the conical tent.

The curved surface area of the tent  

Hence, the canvas required = 
Cost of the canvas at the rate of Rs. 120 per m2 = Rs. 120 × 565.2 = Rs. 67824

Ex 9. How many metres of cloth of 1.1 m width will be required to make a conical tent whose vertical height is 12 m and base radius is 16 m? Find also the cost of the cloth used at the rate of Rs.14 per metre.
 

Sol. Here, h = 12 m, r = 16 m
 = 20m
 Curved surface area =

Width of cloth = 1.1 m

 Cost of the cloth used @ Rs.14 per metre =  × 14 = Rs. 12800.

Ex 10. The curved surface of a right circular cone is 198 cm2 and the radius of its base is 7 cm. Find the volume of the cone.
 

Sol. Radius of the base of the cone = 7 cm.
Let h cm be the vertical height and  cm be the slant height. Here, r = 7 cm.

Now,   81 – 49 = 32 cm

 × 1.41 cm = 5.64 cm

Hence, the volume of the cone = 289.52 cm³.